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# How many millilitres of 0.5 M ${H_2}S{O_4}$ are needed to dissolve 0.5 g of copper (II) carbonate ?

Last updated date: 20th Jun 2024
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Hint: This is a neutralization reaction where the sulphuric acid reacts with copper (II) carbonate to give salt and water. Using the normality equation and then putting the subsequent values, we can get our answer.

This question has sulphuric acid as acid and copper (II) carbonate as the base. There is a neutralization reaction occurring in which the sulphuric acid will react with copper (II) carbonate to form salt and water. We have to calculate the amount of sulphuric acid required. This can be calculated by using normality equation.
Before solving, let us first write the things given to us.
Given :
Molarity of ${H_2}S{O_4}$= 0.5 M
Mass of copper (II) carbonate = 0.5 g
From the given mass of copper (II) carbonate, let us first calculate its normality of copper (II) carbonate.
Normality of copper (II) carbonate = $\dfrac{{0.5 \times 2}}{{123.5}}$ N
Normality of copper (II) carbonate = 0.00809 N

Let ${N_1}$ be the normality of sulphuric acid
${N_2}$ be the normality of copper (II) carbonate
${V_1}$ be the volume of sulphuric acid
${V_2}$ be the volume of copper (II) carbonate = 1000 mL

So, the normality equation is :
${N_1}$${V_1}={N_2}$${V_2}$
Putting the values in equation, we get
$1.0 \times {V_1}$= 0.00809$\times$1000
Solving the equation, we get
${V_1}$ = 8.09 mL
Thus, the 8.09 mL of sulphuric acid is required for 5 g of copper (II) carbonate.

Note: It must be noted that 1 Litre = 1000 mL. We initially suppose the volume of copper (II) carbonate to be 1 L or 1000 mL solution in which the sulphuric acid has to react. Further, always move step by step to decrease chances of mistakes in calculation.